This reaction is most likely to fall under SN2 because the
thing called carbonication does not occur in SN1. The carbon forms a partial
bond with the nucleophile during the intermediate phase and the leaving group.
So for this question the reaction will fall under SN2.
It’s D. Rutherford discovered the atomic nucleus.
This picture represents electrons
Half reaction 1: 2Fe° → Fe₂³⁺ + 6e⁻ /×2.
4Fe° → 2Fe₂³⁺ + 12e⁻.
Iron is oxidized from neutral charge (0) to oxidation number +3, one iron lose three electrons, two irons lose six electrons and four irons twelve electrons.
Half reaction 2: 12e⁻ + 3O₂ → 2O₃²⁻
Oxygen is reduced from neutral chage to oxidation number -2, one oxygen gain two electrons, six oxygens gain twelve electrons.
Balanced chemical reaction: 4Fe + 3O₂ → 2Fe₂O₃.
Answer:
Ca = 10,63
Explanation: Let's see, these are equations, now Look at ca and ca equates to all chemical equations= 10,63
